Nuclear Radius

Probing Matter

Homework

Why are spent fuel rods more radioactive after removal from the reactor?

What are the different types of radioactive waste?

How is each type stored/disposed?

Filling the Gaps

γ emitters can indeed be used as medical tracers as we thought.

Cell membranes can also be destroyed as well as affecting DNA.

Learning ObjectivesLearning Objectives

Explain how an estimate for the Explain how an estimate for the nuclear radius can be obtained from nuclear radius can be obtained from Rutherford’s Experiment.Rutherford’s Experiment.

Rutherford’s Alpha Scattering

At P, the point of closest approach, all of the initial kinetic energy of the alpha is converted to electrostatic potential energy.

Electrostatic Potential Energy

The equation for Electrostatic Potential Energy is given by this equation:-

Where:-•EP is the Electrostatic Potential Energy •rc is the distance of closest approach•ε0 is the permittivity of free space (constant - see data booklet)•Q1Q2 are the charges of the two particles involved.

cP r

QQE 21

04

1

Little bit of Maths… Solving to find rC, what do we get?

Rearranging…

Remember: Q1=2e = 2 × 1.60 × 10-19 C (α is He nucleus) Q2=79e = 79 × 1.60 × 10-19 C (79 protons in Gold

nucleus) EP=7.68 MeV = 768 × 106 × 1.60 × 10-19 J (K.E. of α

particles fired at the foil.) ε0 = 8.85 × 10-12 Fm-1 (from the data booklet)

rc= 2.96 × 10-14 m (a bit large)

Pc E

QQr

0

21

4

A couple of points… The nucleus is treated as a point charge.  At

this level it is not. The alpha particles are stopped some

distance away from the nucleus. It takes higher energy alpha particles to

penetrate the nucleus. The values for the nuclear radius given by

other particles such as protons, neutrons and electrons are slightly different.

…so really it is only an upper limit on the nuclear radius.

Method 2Method 2 About 1 in 10,000 particles are deflected by About 1 in 10,000 particles are deflected by

more than 90º.more than 90º. For a thin foil (so that only one scattering) For a thin foil (so that only one scattering)

with n layers of atoms, the probability of with n layers of atoms, the probability of being deflected is about 1 in 10,000n.being deflected is about 1 in 10,000n.

This probability depends on effective cross This probability depends on effective cross section of nucleus to the atom:-section of nucleus to the atom:-

Typically n=10,000 so d = D/10,000Typically n=10,000 so d = D/10,000nD

d

R

r

000,10

12

41

241

2

2

n

Dd

000,10

22

QuestionsQuestions

A) For a metal foil which has layers of A) For a metal foil which has layers of atoms, explain why the probability of an atoms, explain why the probability of an αα particle being deflected by a given atom is particle being deflected by a given atom is therefore about 1 in 10,000n. (assume 1 in therefore about 1 in 10,000n. (assume 1 in 10000 deflected by more than 90º)10000 deflected by more than 90º)

B) Assuming this probability is equal to the B) Assuming this probability is equal to the ratio of cross sectional area of the nucleus ratio of cross sectional area of the nucleus to that of the atom, estimate the diameter to that of the atom, estimate the diameter of a nucleus for atoms of diameter 0.5 nm of a nucleus for atoms of diameter 0.5 nm in a metal foil of thickness 10 in a metal foil of thickness 10 μμm.m.